Question

A stone falls from a building and reaches the ground in 5 seconds later. how high is the building? (Take g= 10m/s2) *​

Answer 1

Explanation:

Initial velocity, u = 0 m/s

Acceleration due to gravity, g=9.8 m/s

2

Time taken to reach the ground, t = 2.5 sec

Height, h = ?

Using relation,

s=u t+

2

1

gt

2

s=0×2.5+

2

1

×9.8×2.5×2.5

s=0+4.9×2.5×2.5

s = 30.625 m

Answer 2

AnswEr :

GivEn that, a stone falls from a building & reaches the ground in 5 seconds later. And we've to find that, how high is the building. [Height]

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• Initial Velocity, u = 0 m/s.

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• Acceleration due to gravity, g = 9.8 m/s².

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• Time taken by the stone to reach the ground , t = 5 seconds.

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• Height, h = ?

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Directly, we can use the kinematic equation to find the height of the building.

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By using the kinematic Equation :

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$:\implies\sf\;S = ut + \dfrac{1}{2}gt^2$

$:\implies\sf\;S = 0 + \dfrac{1}{2} \times 9.8 \times {5}^{2}$

$:\implies\sf\;S = 0 + \dfrac{1}{2} \times 9.8 \times 10$

$:\implies\sf\;S = 9.8 \times 5$

$\large:\implies\sf\;{ \boxed{ \green{ \sf{S = 49 \: m}}}}$

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$\therefore$ Building is 49 m high.

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⠀Additional information :

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Basically, there are 3 equations of motion. They're as follows :

$\: \: \: \: \: \: \: \: \: \: \bullet\sf\;v = u + at$

$\: \: \: \: \: \: \: \: \: \:\bullet\sf\;v^2 = u^2 + 2as$

$\: \: \: \: \: \: \: \: \: \:\bullet\sf\;S = ut + \dfrac{1}{2}gt^2$